The average of 5 different numbers is 10. What is the highest possible value that one of the numbers can be?

Answer 1

46. If all the other numbers are 1.

Assuming, of course, that all the numbers must be positive, non-zero integers.

1+1+1+1+46 = 50
50 / 5 = 10

Answer 2

The highest value of one of them has to be 19.

To get the highest number, the other four have to have the lowest possible value. That is 0,1,2,and 3. The sum of these is 6.

Let the 5th number be x. Then (6 + x)/5 = 5.

6 + x = 25

x = 25 - 6

x = 19

Check: (0 + 1 + 2 + 3 + 19)/ 5 = 25/5 = 5 OK

Note, the solution requires all of the numbers to be positive and whole. Also, as stated in your problem the five numbers are different.

Answer 3

If the numbers are to be different positive integers,
then the lowest four numbers will be 1, 2, 3 and 4.
This means the fifth number will be as high as possible.

If the fifth number is x, then we know that :
(1 + 2 + 3 + 4 + x) / 5 = 10

That is : (10 + x) / 5 = 10
or, 10 + x = 50
so, x = 40.

Therefore, the highest number must be 40.

Answer 4

infinity.... simple.... If the average is 10 that means that how uch ever large the number can be v can ahve a -ve number nullyfying the effect..

So teh answer goes as infinity...

But the question can a Diffent turn if u say that the they all have to natural numbers in that case the aswer woud be 40 because the greatest values the numbers can take are 1,2,3,4 that would make it a sum of ten but the sum of all the five numbers has to 50 therefore the value of the fifth number has to 40.

Answer 5

the highest value would be 46... the all the other number would be 1....

46+1+1+1+1=50

50/5 = 10

considering it has to be above 0

Answer 6

the average of 5 numbers=10
total of the ten numbers=50
assuming the numbers are natural numbers
the lowest nos can be 1,2,3,4
toatl=1+2+3+4=10
subtracting the fifth number
=50-10=40

Answer 7

un predictable.

(a+b+c+d+e)/5 = 10
a+b+c+d+e = 50
let's assume that "a" be the max of the obove five variables,
then a = 50-(b+c+d+e)

As there is no rule to predict the absolute value of b,c,d and e, the value that "a" can contain to satisfy the abov equation is also un-predictable.


Scenario 1:
Let us consider the following values for b,c,d,and e:
b=infinity, c=one unit lesser than infinity, d=one unit lesser than c, and e=one unit lesser than d
let x = inf,
then a = 50-(x+(x-1)+(x-2)+(x-3))
=50 -(4x-6)
=56 - 4x (4*inf = inf)
=56-inf
= - infinity (anything - infinity is infinity)

Scenario 2:
if b=-infinity, and c=b+1, d=c+1, and e=d+1
then a= 50-(b+c+d+e)
= 50-(-x+(-x+1)+(-x+2)+(-x+3)
= 50+4x-6
= 44+inf (4*inf = inf)
= infinity

Answer 8

infinity... if the other four numbers are negative infinities. but if none of the numbers are negative, then 50 is the highest possible value of one of the numbers. That happens when the other four numbers are zero

Answer 9

Let nos. be a,b,c,d,e
(a+b+c+d+e)/5=10

=>a+b+c+d+e=50
You need to tell if numbers are real, natural or whole

If real the max cannot be determined(Nos. can be infinitely small)

If whole numbers then highest value will be:44
other 4 nos. being 0,1,2,3

If natural numbers then maximum value will be 40 other 4 nos. being 1,2,3,4

Answer 10

Riddiculous question as one of the numbers could be a very low minus number... [i.e. -25,000,000,000,000,000,000,000] Do you follow?